AQA A-Level Organic Chemistry 3.3.6 Organic Analysis Infrared Spectroscopy
3.3.6

Infrared Spectroscopy

Identifying organic functional groups by their molecular vibrations.

Infrared (IR) spectroscopy is a powerful analytical technique used to determine the functional groups present in organic molecules. By exposing a compound to infrared radiation, we can observe which frequencies are absorbed by its chemical bonds.

🔑 Key Principle 1

Covalent bonds are not rigid, they behave like springs. They vibrate naturally by stretching or bending. Each covalent bond absorbs infrared radiation of a specific frequency that matches its natural vibration frequency. This absorption results in characteristic dips (troughs) on an IR spectrum.

🔑 Key Principle 2

Greenhouse gases in the atmosphere, such as carbon dioxide, water vapour, and methane, absorb infrared radiation emitted by the Earth. This absorption causes their covalent bonds to vibrate more vigorously, re-radiating heat back towards the Earth and warming the atmosphere.

How Infrared Spectroscopy Works

When infrared light is passed through an organic sample, bonds of different types (such as \( \text{C-H} \), \( \text{C}=\text{O} \), or \( \text{O-H} \)) absorb specific wavelengths of energy. An IR spectrum plot shows Transmittance (%) on the y-axis against Wavenumber (\( \text{cm}^{-1} \)) on the x-axis.

Wavenumber is defined as \( 1/\lambda \) (where \( \lambda \) is the wavelength). It is directly proportional to frequency and energy. As bonds absorb energy, the amount of light transmitted drops, producing downward-pointing peaks.

Key Absorptions Table

You will be provided with these values in your AQA exam formula booklet, but you must know how to apply them to identify compounds:

Bond Functional Group / Type Wavenumber Range (\( \text{cm}^{-1} \)) Appearance of Peak
O-H Alcohols 3230-3550 Broad, smooth, and rounded
O-H Carboxylic acids 2500-3300 Very broad, distorted, overlapping C-H
C=O Aldehydes, Ketones, Esters, Acids 1680-1750 Strong and sharp
C-O Alcohols, Esters, Carboxylic acids 1000-1300 Strong (in fingerprint region)
N-H Amines, Amides 3300-3500 Sharp, often with multiple spikes
C-H Alkanes, Alkenes, Arenes 2850-3100 Sharp (present in almost all organic compounds)
Infrared Spectroscopy

An analytical technique used to identify functional groups in organic molecules by measuring the absorption of infrared radiation by vibrating covalent bonds.

Fingerprint Region

The complex region of an infrared spectrum below 1500 \( \text{cm}^{-1} \). It contains many overlapping peaks that are unique to each individual compound, allowing for exact identification when compared against a library of known spectra.

Wavenumber

The reciprocal of wavelength, measured in \( \text{cm}^{-1} \). It is the standard unit of frequency used in infrared spectroscopy.

Greenhouse Gas

A gas in the atmosphere (such as \( \text{CO}_2 \), \( \text{H}_2\text{O} \), or \( \text{CH}_4 \)) that absorbs infrared radiation emitted from the Earth's surface and re-radiates it, trapping heat in the atmosphere.

Interpreting an IR Spectrum

To identify an organic compound, look for prominent peaks outside the fingerprint region (usually above 1500 \( \text{cm}^{-1} \)):

Annotated Carboxylic Acid IR Spectrum Wavenumber (cm⁻¹) Transmittance (%) 4000 3500 3000 2500 2000 1500 1000 500 100 50 0 Fingerprint Region (below 1500 cm⁻¹) Broad O-H stretch (2500-3300 cm⁻¹) Sharp C=O stretch (1680-1750 cm⁻¹)
📝 AQA Examiner Tip

Be very careful when distinguishing between the O-H absorption in an alcohol and the O-H absorption in a carboxylic acid. The O-H in an alcohol is a broad, smooth band located at 3230-3550 \( \text{cm}^{-1} \). The O-H in a carboxylic acid is much broader and more distorted, spanning 2500-3300 \( \text{cm}^{-1} \), and it always overlaps with the C-H stretching peaks around 2950 \( \text{cm}^{-1} \).

The Fingerprint Region

The region of the spectrum below 1500 \( \text{cm}^{-1} \) is known as the fingerprint region. It contains a complex pattern of peaks unique to each specific organic compound, arising from the bending vibrations of the whole molecule.

In examinations, you do not need to identify individual peaks in the fingerprint region. Instead, you must know that it is used to identify a compound by comparing the spectrum against a library of known spectra of pure compounds. If the fingerprint region of your sample matches a database spectrum exactly, it confirms the compound's identity.

Greenhouse Gases and Global Warming

The Earth absorbs solar radiation and re-emits it as longer-wavelength infrared radiation. Atmospheric gases such as carbon dioxide (\( \text{CO}_2 \)), methane (\( \text{CH}_4 \)), and water vapour (\( \text{H}_2\text{O} \)) contain polar covalent bonds.

These bonds absorb this infrared radiation, matching their natural vibrational frequencies. This absorption causes the bonds to vibrate more, trapping thermal energy in the atmosphere. The concentration of these greenhouse gases determines how much thermal energy is trapped, direct links exist between increased emissions and global warming.

📝 AQA Examiner Tip

AQA can ask why certain molecules act as greenhouse gases. Your answer must state that the molecules contain polar bonds that absorb infrared radiation emitted by the Earth, causing the bonds to vibrate more vigorously and trap heat.

Worked Examples

✏️ Worked Example 1
An unknown organic compound has the molecular formula \( \text{C}_3\text{H}_6\text{O} \). Its infrared spectrum shows a strong, sharp absorption peak at 1715 \( \text{cm}^{-1} \) but shows no absorption bands above 3000 \( \text{cm}^{-1} \). Deduce the functional group present and propose a structure for the compound.

Solution:

1. The molecular formula \( \text{C}_3\text{H}_6\text{O} \) indicates a degree of unsaturation of 1 (containing either a ring or a double bond).

2. The strong, sharp peak at 1715 \( \text{cm}^{-1} \) corresponds to a carbonyl group (\( \text{C}=\text{O} \)). This accounts for the degree of unsaturation.

3. The absence of any broad peak above 3000 \( \text{cm}^{-1} \) confirms the absence of an \( \text{O-H} \) group. This rules out alcohols and carboxylic acids.

4. Since the compound contains a carbonyl group and no hydroxyl group, it must be either an aldehyde or a ketone. With three carbon atoms, the possible structures are propanal, \( \text{CH}_3\text{CH}_2\text{CHO} \), or propanone, \( \text{CH}_3\text{COCH}_3 \). Both match the infrared data given.

✏️ Worked Example 2
Explain how you would use infrared spectroscopy to distinguish between structural isomers propanoic acid and methyl ethanoate. State the key differences in their spectra.

Solution:

1. Identify functional groups: Propanoic acid is a carboxylic acid, containing both a carbonyl group (\( \text{C}=\text{O} \)) and an acidic hydroxyl group (\( \text{O-H} \)). Methyl ethanoate is an ester, containing a carbonyl group (\( \text{C}=\text{O} \)) and a \( \text{C-O} \) single bond, but no \( \text{O-H} \) group.

2. Carboxylic acid spectrum features: The spectrum of propanoic acid will display a very broad absorption band between 2500-3300 \( \text{cm}^{-1} \), representing the \( \text{O-H} \) stretch. It will also show a sharp carbonyl peak at 1680-1750 \( \text{cm}^{-1} \).

3. Ester spectrum features: The spectrum of methyl ethanoate will display a sharp carbonyl peak at 1680-1750 \( \text{cm}^{-1} \) but will have no broad absorption band in the 2500-3300 \( \text{cm}^{-1} \) region (no \( \text{O-H} \) bond).

4. Exact matching: The fingerprint regions below 1500 \( \text{cm}^{-1} \) of both compounds can be compared against a database of known spectra to confirm the exact identity of each isomer.

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